You have found the following ages (in years) of 5 zebras. The zebras are randomly selected from the 37 zebras at your local zoo: $ 13,\enspace 1,\enspace 10,\enspace 37,\enspace 7$ Based on your sample, what is the average age of the zebras? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 37 zebras, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{0.36} + {158.76} + {12.96} + {547.56} + {43.56}} {{5 - 1}} $ $ {s^2} = \dfrac{{763.2}}{{4}} = {190.8\text{ years}^2} $ We can estimate that the average zebra at the zoo is 13.6 years old. There is a variance of 190.8 years $^2$.